All categories

3 years ago

CAN SOMEONE PLEASE HELP ME

Physics

2 answers:

VLD [36.1K]3 years ago

7 0

Answer:

she will eventually slow down and come to a stop

lakkis [162]3 years ago

7 0

2: she will eventually slow down and come to a stop

You might be interested in

LO NECESITO PARA HOY URGENTE!! 2. Exprese en metros las siguientes longitudes a) 48,9 km b) 36,875 cm c) 756,34 hm d) 9876 mm

Vesnalui [34]

Answer:

a)48900 metros

b)0.36875 metros

c)75634 metros

d)9.876 metros

Explanation:

Hola, para resolver debemos convertir unidades utilizando equivalencias

a) 48.9 km

1 kilometro = 1000 metros

48.9 x 1000 = 48900 metros

b) 36.875 cm

1 centímetro =0.01 metros

36.875 x 0.01 = 0.36875 metros

c) 756,34 hm

1 hectómetro= 100 metros

756.34 x 100 = 75634 metros

d) 9876 mm

1 milímetro = 0.001 m

9876 x 0.001 = 9.876 metros

4 0

3 years ago

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

daser333 [38]

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

$T = FR = 18 * 2.4 = 43.2 Nm$

According to Newton 2nd law, the angular acceleration would be

$\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2$

It starts from rest, then after 15s it would achieve a speed of

$\omega = \alpha t = 0.021 * 15 = 0.31 rad/s$

(b) The distance angle swept by it is:

$\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad$

Hence the work by the child

$W = T\theta = 43.2 *2.314 \approx 100 J$

c) Average power to work per time unit

$P = \frac{W}{t} = \frac{100}{15} = 6.67 W$

7 0

3 years ago

A geologic event causes changes to the physical makeup of a particular place and occurs _____.

babymother [125]

A geologic event causes changes to the physical makeup of a particular place and occurs slowly.

Geological events are what causes numerous changes and phenomena on the Earth's surface. Examples of these events include cliff erosion, volcanic eruption, or sedimentation at a mouth of a river.

Geological processes are extremely slow. However, because of the immense lengths of time involved, huge physical changes do occur - mountains are created and destroyed, continents form, break up and move over the surface of the Earth, coastlines change and rivers and glaciers erode huge valleys.

Geological events are both classified as internal and external. This means that these events occur both in the Earth's surface and interior.

4 0

3 years ago

Read 2 more answers

When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil

Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

$2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}$

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

$2t = 2\lambda_{film}$

$t = \lambda_{film}$

The index of refraction of soap is given, then

$\lambda_{film} = \frac{\lambda_{vacuum}}{n}$

Combining the results of all steps we get

$t = \frac{\lambda_{vacuum}}{n}$

Rearranging, we find

$\lambda_{vacuum} = tn$

$\lambda_{vacuum} = (278)(1.33)$

$\lambda_{vacuum} = 369.74nm$

4 0

3 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

4 years ago